Prove an upper bound for the determinant of a matrix A 
*

*Let $A$ be a $3 \times 3$ real matrix with all $0\le a_{ij} \le 1$.
Show that $\det(A) \leq 2$ and find such matrices with $\det(A) = 2$.


*Let $A$ be a $n \times n$ matrix with all $0\le a_{ij} \le 1$.
Estimate precisely a maximum possible value of $\det(A)$.
I would like to try and solve this problem without the usage of the permutation formula for the determinant.
 A: Edit 3: This hint might not be the best way to approach the problem.
Hint: Start with $2\times 2$ matrices and work by induction.
Edit 2: 
Show that for a $2\times 2$ matrix with those properties, $|\det A| \leq 1$. Evaluate different cases and convince yourself these are in fact the extrema ($-1$ and $1$). The matrices that achieve these values are:
$$
\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix},\quad \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix},\quad\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix},
$$
and
$$
\begin{pmatrix} 0 & 1 \\ 1 &0\end{pmatrix},\quad \begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix},\quad\begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix}.
$$
Now, if $A$ is your  $3\times 3$ we have $\det A = a_{11}\det A_{11} - a_{22}\det A_{22} + a_{33}\det A_{33}$. Can you find the matrices such that $|\det A| = 2$ ? ATM I don't know how to prove this is in fact the maximum.
A: @hjhjhj57, not sure if you've found the matrices yet - the ones where detA = 2.  But here is my work.  If you just play around with the 3 cases, and see which entries must be zero and which entries must be equal to one, you should get the below 3 matrices - and also using your hint for induction on the 1x1 and 2x2 cases.  
Look at:
$$A=
        \begin{bmatrix}
        a & b & c \\
        d & e & f \\
        g & h & i \\
        \end{bmatrix}
$$
$\implies$ detA = a(ei-fh) - b(di - fg) + c(dh-eg).
Call the first term I, the second term II, and the third term III.
Case 1:
Now if we want to maximize detA, by maximizing I and II (so, making b(di-fg) = -1), then from how the variables are related and already chosen, you'll see that III will be negative if c were not equal to zero.  So we set c equal to zero to maximize detA.
Case 2:
If we want to maximize detA, by maximizing II and III - so making II and III each equal to 1 - then we'll see that I is a negative number.  So we must set a = 0 to maximize detA.
Case 3:
Similar argument for maximizing I and III.
In all 3 cases, the maximum determinant was equal to 2, so we see that there are exactly 3 matrices with max det(A) = 2, which are:
$$
        \begin{bmatrix}
        1 & 1 & 0 \\
        0 & 1 & 1 \\
        1 & 0 & 1 \\
        \end{bmatrix}
$$
$$
        \begin{bmatrix}
        0 & 1 & 1 \\
        1 & 0 & 1 \\
        1 & 1 & 0 \\
        \end{bmatrix}
$$
$$
        \begin{bmatrix}
        1 & 0 & 1 \\
        1 & 1 & 0 \\
        0 & 1 & 1 \\
        \end{bmatrix}
$$
